| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Find sample size for test |
| Difficulty | Standard +0.3 This is a standard hypothesis test for a binomial proportion with straightforward application of the normal approximation or binomial tables. Part (a) follows a routine procedure (state hypotheses, calculate test statistic, compare to critical value), while part (b) requires working backwards to find a critical value, which is slightly less routine but still a textbook exercise requiring no novel insight. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(X \sim B(50, p)\) | B1 B1 | |
| \(H_0 : p = 0 \cdot 1\), \(H_1 : p > 0 \cdot 1\) | M1 A1 A1 | |
| Under \(H_0\), \(P(X \geq 9) = 1 - P(X \leq 8) = 1 - 0.9421 = 0.0579 > 5\%\), so do not reject \(H_0\) | M1 A1 A1 | |
| (b) Need \(P(X \geq n) < 0.01\), so \(n = 11\) Need 11 faulty | M1 M1 A1 | Total: 9 marks |
(a) $X \sim B(50, p)$ | B1 B1 |
$H_0 : p = 0 \cdot 1$, $H_1 : p > 0 \cdot 1$ | M1 A1 A1 |
Under $H_0$, $P(X \geq 9) = 1 - P(X \leq 8) = 1 - 0.9421 = 0.0579 > 5\%$, so do not reject $H_0$ | M1 A1 A1 |
(b) Need $P(X \geq n) < 0.01$, so $n = 11$ Need 11 faulty | M1 M1 A1 | **Total: 9 marks**A supplier of widgets claims that only 10\% of his widgets have faults.
\begin{enumerate}[label=(\alph*)]
\item In a consignment of 50 widgets, 9 are faulty. Test, at the 5\% significance level, whether
this suggests that the supplier's claim is false. [6 marks]
\item Find how many faulty widgets would be needed to provide evidence against the claim at
the 1\% significance level. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q4 [9]}}